\appendix{定理\ref{Thm:IC:MISO:Centralized:SDP:Rank1}的证明}
\label{Apx:IC:MISO:Centralized:SDP:Rank1}
问题\eqref{Eqn:IC:MISO:Centralized:SDP}的拉格朗日函数为
\begin{align}
&L\left( {\left\{ {{{\bf{W}}_i}} \right\}} \right) = \sum\limits_{i = 1}^K {{\varepsilon _i}{\rm{Tr}}\left( {{{\bf{W}}_i}} \right)}  - \sum\limits_{i = 1}^K {{\rm{Tr}}\left( {{{\bf{X}}_i}{{\bf{W}}_i}} \right)}  + \sum\limits_{i = 1}^K {{\tau _i}{c_i}} \nonumber\\
&- \sum\limits_{i = 1}^K {{\tau _i}\left( {{u_i}{\rm{Tr}}\left( {{{\bf{W}}_i}{{\bf{g}}_{i,i}}{\bf{g}}_{i,i}^T} \right) - \sum\limits_{j = 1,j \ne i}^K {{\lambda _{i,j}}{\rm{Tr}}\left( {{{\bf{W}}_j}{{\bf{g}}_{i,j}}{\bf{g}}_{i,j}^T} \right)} } \right)}\nonumber\\
& { + \sum\limits_{n = 1}^N {{z_n}\left( {{\rm{Tr}}\left( {{\bf{W}}_i{{\bf{e}}_n}{\bf{e}}_n^T} \right) - \frac{{A_i^2}}{{b_n^2}}} \right)} }.
\end{align}
式中，
${{\tau _i}}$，${{z_n}} $和${\mathbf{X}_i} \in {\mathbb R}^{{N}}$ are
分别为约束 \eqref{Eqn:IC:MISO:Centralized:SDP:b}， \eqref{Eqn:IC:MISO:Centralized:SDP:c}和 \eqref{Eqn:IC:MISO:Centralized:SDP:d}的拉格朗日对偶变量；
${{\tau _i}}\geq 0$，${{z_n}} \geq 0$，${\mathbf{X}_i} \succeq {\mathbf{0}}$.

问题\eqref{Eqn:IC:MISO:Centralized:SDP}的KKT条件如下所示
\begin{subequations}
    \begin{align}
    &{\varepsilon _i}{\bf{I}} - {{\bf{X}}_i} - {\tau _i}{u_i}{{\bf{g}}_{i,i}}{\bf{g}}_{i,i}^T + \sum\limits_{j = 1,j \ne i}^K {{\tau _i}{\lambda _{ji}}{{\bf{g}}_{j,i}}{\bf{g}}_{j,i}^T}+\sum\limits_{n = 1}^N {{z_n}{{\bf{e}}_n}{\bf{e}}_n^T}  = {\bf{0}},\forall i \in \mathcal{K},\label{Apx:IC:MISO:Centralized:SDP:Rank1:KKT:a} \\
    &{\tau _i}\left( {{c_i} - {u_i}{\rm{Tr}}\left( {{{\bf{W}}_i}{{\bf{g}}_{i,i}}{\bf{g}}_{i,i}^T} \right) + \sum\limits_{j = 1,j \ne i}^K {\lambda _{i,j}}{{\rm{Tr}}\left( {{{\bf{W}}_j}{{\bf{g}}_{i,j}}{\bf{g}}_{i,j}^T} \right)} } \right)= 0, \forall i \in \mathcal{K},\label{Apx:IC:MISO:Centralized:SDP:Rank1:KKT:b} \\
    &{z_i}\left( {{\rm{Tr}}\left( {{\bf{W}}{{\bf{e}}_i}{\bf{e}}_i^T} \right) - \frac{{{b_i}}}{A}} \right) = 0,\forall i \in \mathcal{K},\label{Apx:IC:MISO:Centralized:SDP:Rank1:KKT:c} \\
    &{{\bf{X}}_i}{{\bf{W}}_i} = {\bf{0}},\forall i \in \mathcal{K},\label{Apx:IC:MISO:Centralized:SDP:Rank1:KKT:d} \\
    &{u_i}{\rm{Tr}}\left( {{{\bf{W}}_i}{{\bf{g}}_{i,i}}{\bf{g}}_{i,i}^T} \right) \ge \sum\limits_{j = 1,j \ne i}^K {{\lambda _{i,j}}{\rm{Tr}}\left( {{{\bf{W}}_j}{{\bf{g}}_{i,j}}{\bf{g}}_{i,j}^T} \right)}  + {c_i},\forall i \in \mathcal{K},\label{Apx:IC:MISO:Centralized:SDP:Rank1:KKT:e} \\
    &{\rm{Tr}}\left( {{{\bf{W}}_i}{{\bf{e}}_n}{\bf{e}}_n^T} \right) \le \frac{{b_i^2}}{{A_i^2}},\forall n \in \mathcal{N},\forall i \in \mathcal{K},\label{Apx:IC:MISO:Centralized:SDP:Rank1:KKT:f} \\
    &{{\bf{W}}_i} \succeq {\bf{0}},{{\bf{X}}_i} \succeq {\bf{0}},i \in \mathcal{K},{\tau _i} \ge 0,{z_i} \ge 0,\forall i \in \mathcal{K}.\label{Apx:IC:MISO:Centralized:SDP:Rank1:KKT:g}
    \end{align}
\end{subequations}

通常情况下，问题\eqref{Eqn:IC:MISO:Centralized:SDP}是满足Slater条件。如果问题\eqref{Eqn:IC:MISO:Centralized:SDP}是可达的， 根据 问题\eqref{Eqn:IC:MISO:Centralized:SDP}的结构，它一定存在一个严格可行解。因此，强对偶性成立，并且如果${\mathbf{W}}$，${\mathbf{X}}$，$\left\{ {{x_k}} \right\}$，$\left\{ {{y_i}} \right\}$是原始-对偶问题的最优解，那么他们应当满足上述KKT条件。


将\eqref{Apx:IC:MISO:Centralized:SDP:Rank1:KKT:a}改写为
\begin{align}
{{\bf{X}}_i} = &{{\varepsilon _i}}{\bf{I}} - {\tau _i}{u_i}{{\bf{g}}_{i,i}}{\bf{g}}_{i,i}^T + \sum\limits_{j = 1,j \ne i}^K {{\tau _j}{\lambda _{ji}}{{\bf{g}}_{j,i}}{\bf{g}}_{j,i}^T}+ \sum\limits_{n = 1}^N {{z_n}{{\bf{e}}_n}{\bf{e}}_n^T}, \forall i \in \mathcal{K}.  \label{Apx:IC:MISO:Centralized:SDP:Rank1:KKT:a_eq}
\end{align}



假设$ s_j $服从第二章给出的ABG分布，则它的微分熵$h\left( {{s_j}} \right)$可由下式计算得到
\begin{align}\label{Apx:IC:MISO:Centralized:SDP:Rank1Entropy:a}
&h\left( {{s_j}} \right) = \frac{{1{\rm{ + }}{\alpha _j} + {\gamma _j}{\varepsilon _j}}}{{{\rm{ln}}2}}.
\end{align}
此外，因为$h\left( Q \right) \le \frac{1}{2}\log 2\pi e{\mathop{\rm var}} \left( Q \right)$对于任意一个方差为${\mathop{\rm var}} \left( Q \right)$的随机变量，所以存在如下关系
\begin{align}\label{Apx:IC:MISO:Centralized:SDP:Rank1Entropy:b}
h\left( {{s_j}} \right) \le \frac{1}{2}{\log _2}2\pi e{\varepsilon _j}.
\end{align}
联立\eqref{Apx:IC:MISO:Centralized:SDP:Rank1Entropy:a}和 \eqref{Apx:IC:MISO:Centralized:SDP:Rank1Entropy:b}，可以得到
\begin{align}
\pi e{\varepsilon _j} \ge {e^{1{\rm{ + }}2\left( {{\alpha _j} + {\gamma _j}{\varepsilon _j}} \right)}}.
\end{align}
因此，${\lambda _{i,j}} = 2\pi {\varepsilon _j}{2^{2{R_i}}} - {e^{1 + 2\left( {{\alpha _j} + {\gamma _j}{\varepsilon _j}} \right)}} \ge 0$.


因为${{\varepsilon _i}}{\bf{I}} + \sum\limits_{j = 1,j \ne i}^K {{\tau _j}{\lambda _{ji}}{{\bf{g}}_{j,i}}{\bf{g}}_{j,i}^T}  + \sum\limits_{n = 1}^N {{z_n}{{\bf{e}}_n}{\bf{e}}_n^T}  \succ {\bf{0}}$，并且
${\text{rank}}\left({\tau _i}{u_i}{{\bf{g}}_{i,i}}{\bf{g}}_{i,i}^T \right) = 1$，所以${\text{rank}}\left( {\mathbf{X}_i} \right) \geq  {N} - 1$ 。

此外，又由于${{\bf{X}}_i}{{\bf{W}}_i} = {\bf{0}}$，故${\text{rank}}\left( {\mathbf{W}_i} \right) \leq 1$。如果${\text{rank}}\left( {\mathbf{W}_i} \right) = 0$，则可以解得${\mathbf{W}_i} = {\mathbf{0}}$，但是这会导致当$R_i > 0$时，约束\eqref{Eqn:IC:MISO:Centralized:SDP:b}无法被满足。


因此， ${\text{rank}}\left( {\mathbf{W}_i} \right) =1$。
